Extraneous root of fractional equation
What is the sum of all the integers a such that the following equation has no real roots:
x − 1 x 2 + x + a − 5 = a ?
The answer is 18.
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2 solutions
Moderator note:
For those who do not understand why a = 3 has no real roots, try working through this problem:
Solve for real values of x : x − 1 x = x − 1 1 .
You can see the full solution here .
I think there is a problem with that thinking. If a = 3 , then the expression simplifies to x − 1 = 0 , a straight line, which has a real root x = 1 . I think it would create a nightmare if we're to say that simplified expressions are "not equivalent" to the originals. If a function plots like a straight line, looks like a straight line, and simplifies to a linear equation, it nevertheless may not "really" be a straight line, but has a hole in it? As an easy example, what is the limit value of x − 1 x 2 − 1 as x → 1 ? I would say " 2 ", not "indeterminate".
I would say that the wording of this problem is that, given an integer value " a ", what would be the roots of the equation? This means, given that a = 3 , we know it should reduce to a simple linear equation, with x = 1 as the root.
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I disagree with your first line. When a = 3 , the expression simplifies to x + 2 = 0 , x = 1 . In fact, at no point in time can we take x = 1 , since we cannot divide by 0 = x − 1 .
There is a huge difference between "What is the limit of lim x → 1 x − 1 x 2 − 1 " and "What is the value of the function x − 1 x 2 − 1 when x = 1 ".
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Calvin, your original statement: "then there are likely no real roots, since we cannot divide by 0" is the best way of putting it. The truth is that this is an ill defined neighborhood in the domain of the function. It really does not mean there is no real zero. The right way is handle it is to exclude x = 1 from the domain of the function. However, given that the function behaves well around x=1, it is unclear whether there is a real solution or not.
That's why mathematics continues to perplex and surprise, because of stuff like this. If one uses Mathematica, for instance, to solve x − 1 x 2 − 1 − 2 = 0 , it gives x = 1 as the root. And yet if it's used to find the value of x − 1 x 2 − 1 − 2 at x = 1 , it says it's indeterminate. In other words, sometimes if one plugs back into f ( x ) the very root it supposes to have, it doesn't get you f ( x ) = 0 ! This gives me such an headache. (The 0 ! is meant to be a disgusted surprise, not zero factorial 0 ! = 1 ).
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@Michael Mendrin – Wolfram Alpha says that there are no solutions. It does state the alternate form correctly.
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@Calvin Lin – Oh gosh, what are we're supposed to do now when even Mathematica and Wolfram Alpha don't agree? They're both made by the same company! I knew I should have majored in law.
There's another problem somewhat similar to this It's confusing , where the fallacy comes from assuming that expressions are necessarily the same "after simplification".
Oh my god. This problem makes me want to leave algebra😢😟😨😭 Who would have thought about a= 3??
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Same! It killed my rating! :O
I got tricked the first time too. Very nice problem!
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I got tricked too. But, then the title of the problem saved me big time.
This is probably the reason the rating is >1750. I wondered why such a simple looking problem has a level 4 rating.
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exactly.. although it is rather evident that we can't have x=1<=>a=3 in the solution.
Now it has become a level 5.
This shows that blind algebraic manipulation won't help!!....... Thank you for posting such a nice problem
explain in detail pls
wow i even didn't thought upon x-1
what a good explanation...thx calvin
Made the same mistake on the first try but realized my mistake and checked for when x = 1 on my second try.
didnot read the question clearly
CAN'T GET WHY A=3?? SHOULDN'T IT BE ONLY 4,5 AND 6??
If a=3 is a solution, then the corresponding value of x=1. But that makes the expression undefined. What am i missing @Calvin?
can't we say, for a=3, x=1 is the root of the equation which is real. or in other way for x=1 x − 1 x 2 + x − 2 has 0/0 form, which is indeterminate and by L'hospitals rule has got the value 3? I think it is valid thinking.
Yes I agree with your explanation, instead of factorizing ,made it perfect square,
2 5 − 4 − 1 0 a + a 2 = ( a − 5 ) 2 − 4 < 0
Thus,
it clearly shows 3 can't be , at a=3 the parabola will touch the axis. or D=0
OR
( a − 5 ) 2 < 2 2
− 2 < a − 5 < 2 , 3 < a < 7
what's the problem here @Calvin Lin
and who said here the function is onto?
hi i didn't understand after (a-3)(a-7) please clarify
Brilliant.
I disagree with you!!! if we get as a = 3 ,
( x² + x + a - 5 ) / (x- 1) = a
(x²+ x - 2) / ( x- 1) = 3
(x + 2) (x - 1) / (x-1) = 3
x + 2 = 3
so x= 1
BUT IN THE SOLUTION IT SAYS THAT x IS NOT A REAL NUMBER
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When you substitue in x = 1 , you get 0 0 which is undefined (and in particular not equal to 3).
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in the problem it says that it must not contain REAL roots. that means when you solve the QUADRATIC equation you must get a non REAL answer for X.
as you say if i get as, X=1 then X is a real number.so it is not correct to substitute 1 (a real number) as X.
an as you say if i substitute 1 as X then "" a ''''' becomes a NON-REAL NUMBER (complex number) .But in the question it doesn't say that''' a ''' is a Complex number.
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@Hiruja Mahagedara – I do not understand what you are saying. Do you agree with the following:
- "But in the question it doesn't say that a is a complex number": In the question, it is stated that a is required to be an integer
- "It is not correct to substitute 1 for x": I agree that we do not have equality for a = 3 , x = 1 .
This tells us that a = 3 is one of the solutions to this problem, and we have to find the rest.
@Hiruja Mahagedara – I'm really sorry Calvin.....but substituting a=3 we get D=0 and the solution to the equation is x=1
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@Anubhav Mahapatra – Consider this different problem:
What is the solution to x − 1 1 = x − 1 x ? What happens when we substitue x = 1 into the equation?
There is actually no solution to the equation x − 1 1 = x − 1 x . When we try to solve it by clearing the denominators, we must remember that we cannot divide by 0, meaning that x = 1 . We run into a similar issue in this question when a = 3 , namely
x − 1 x 2 + x − 2 = 3 has no real roots, since x = 1 is not a root of the expression.
x − 1 x 2 + x − 2 = 3
Why is 7 not included sir Calvin? The range of a's is 3<x<7 right?
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Oh.. I get it now.. when a=7 x=3 which is a real solution.. Yes, i got 3 but missed and included 7.. hooooooo
Exactly Same Way.
No have value x to a=3 real or not real Equation invalid to a=3
Okay, you did it assuming the the denominator can't be 0. But why did you use it to make the numerator 0? Thanks :)
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I did not "use it to make the numerator 0".
If the denominator is non-zero, then we can multiply throughout by the denominator and get equivalent algebraic statements.
In the event that the denominator is 0, the statement might not be true (or it might still be), so they will not be equivalent algebraic statements.
I forgot about x=1 the first time as well.
I have a question about the step where the numerator is set to 0 when x = 1. Isn't any real number divided by 0 undefined? Why does the numerator have to be 0? Thanks!
Thanks calvin for this nice solution and specially for the explaination for a=3
Calvin when we divide anything by zero it becomes undefined. If this expression becomes undefined how can we find a solution for that. And becomes undefined when x=1 which is a real root which contradicts the question given. Plz give me explanation about this.
I think there is a problem with that thinking. If , then the expression simplifies to , a straight line, which has a real root . I think it would create a nightmare if we're to say that simplified expressions are "not equivalent" to the originals. If a function plots like a straight line, looks like a straight line, and simplifies to a linear equation, it nevertheless may not "really" be a straight line, but has a hole in it? As an easy example, what is the limit value of as ? I would say "", not "indeterminate". I would say that the wording of this problem is that, given an integer value "", what would be the roots of the equation? This means, given that , we know
x=1 really should be excluded from the domain, it is too misleading, and something divided by zero is undefined, not unreal
The integers a are 3 , 4 , 5 , 6 , with a sum of 18. Most people miss the solution a = 3 .
If x − 1 = 0 , then we can multiply throughout to get x 2 + x + a − 5 = a ( x − 1 ) ⇒ x 2 + ( 1 − a ) x + 2 a − 5 . The discriminant is ( 1 − a ) 2 − 4 ( 2 a − 5 ) = 2 1 − 1 0 a + a 2 = ( a − 3 ) ( a − 7 ) . If this is negative, then there are no real roots, which gives us the values of a = 4 , 5 , 6 .
However, if x − 1 = 0 , then there are likely no real roots, since we cannot divide by 0. This occurs when 1 2 + 1 + a − 5 = 0 , or when a = 3 . We can see that x − 1 x 2 + x − 2 = 3 has no real roots, since x = 1 is not a root of the expression.